--- title: "Percentage energy loss of 25 kVA transformer" subject: "Transmission and Distribution System Design" chapter: 7 key: "semester_8/transmission_and_distribution_system_design:transformer-energy-loss-25kva" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/transformer-energy-loss-25kva site: BCT Engineering --- # Percentage energy loss of 25 kVA transformer Transmission and Distribution System Design · Chapter 7: Distribution System Design - Asked in: 2080 Chaitra, 8 marks - Video: [How to Calculate Transformer Losses at a Percent Full Load (PE Power Exam)](https://www.youtube.com/watch?v=5pUQFrbX3hI) — Raiya Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/transformer-energy-loss-25kva - Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-7 ## Question A distribution transformer of 25 kVA with no-load loss of 72 W and rated copper loss is 300 W. Consider the demand for the transformer is 20 kW and the monthly energy sell equals 4800 units @ 0.85 power factors. Use Loss of Load Factor (LLF) = 0.3LF + 0.7LF². Find the percentage energy loss. ## Short answer Load factor = 4800/(20×720) = 0.333, LLF = 0.1778. Peak copper loss = 300×(23.53/25)² = 265.7 W. Monthly core loss 51.84 kWh, copper loss 34.02 kWh. Total loss 85.86 kWh gives percentage loss = 85.86/4800×100 ≈ 1.79%. ## Answer Percentage energy loss = (core energy loss + copper energy loss) / energy sold × 100. Copper energy loss uses the loss of load factor. **Assumption:** month = 30 days = 720 h. #### Step 1: Load factor $$ LF = \frac{\text{Energy}}{P_{max} \times T} = \frac{4800}{20 \times 720} = 0.3333 $$ #### Step 2: Loss of load factor $$ \begin{aligned} LLF &= 0.3\,LF + 0.7\,LF^2 \\ &= 0.3(0.3333) + 0.7(0.3333)^2 \\ &= 0.1000 + 0.0778 = 0.17778 \end{aligned} $$ #### Step 3: Copper loss at peak load $$ \begin{aligned} S_{peak} &= \frac{20}{0.85} = 23.529 \text{ kVA} \\ P_{cu,peak} &= 300 \times \left(\frac{23.529}{25}\right)^2 = 300 \times 0.8858 = 265.74 \text{ W} \end{aligned} $$ #### Step 4: Monthly energy losses $$ \begin{aligned} E_{cu} &= 0.26574 \times 0.17778 \times 720 = 34.02 \text{ kWh} \\ E_{core} &= 0.072 \times 720 = 51.84 \text{ kWh} \\ E_{loss} &= 34.02 + 51.84 = 85.86 \text{ kWh} \end{aligned} $$ #### Step 5: Percentage energy loss $$ \%E_{loss} = \frac{85.86}{4800} \times 100 = 1.79\% $$ (Based on energy input $4800 + 85.86 = 4885.86$ kWh, it is $85.86/4885.86 = 1.76\%$.) **Answer: Percentage energy loss ≈ 1.79% of energy sold** (monthly loss ≈ 85.9 kWh: 51.84 kWh core loss + 34.02 kWh copper loss). Note that core loss is the larger part, because the load factor is low (0.33): the transformer is energised all month but carries peak load only briefly. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With 4800 kWh, 20 kW peak and 720 h, the load factor is: A. 0.50 B. 0.3333 C. 0.8 D. 0.2 Answer: B. LF = 4800/(20 x 720) = 0.3333. (Key point: Load factor) 2. LLF = 0.3LF + 0.7LF^2 for LF = 0.3333 equals: A. 0.1778 B. 0.3333 C. 0.1111 D. 0.2333 Answer: A. 0.1 + 0.0778 = 0.1778. (Key point: LLF) 3. Peak copper loss is found by: A. Using 300 W directly B. 300 x (23.53/25)^2 = 265.7 W, with S = 20/0.85 C. 300 x (20/25)^2 D. 300 x 0.85 Answer: B. Copper loss scales with the square of kVA load; peak is 20/0.85 = 23.53 kVA. (Key point: Copper loss at peak) 4. Total monthly loss and its percentage of energy sold: A. 85.86 kWh, 1.79% B. 51.84 kWh, 1.08% C. 34.02 kWh, 0.7% D. 85.86 kWh, 17.9% Answer: A. Core 51.84 plus copper 34.02 = 85.86 kWh; 85.86/4800 = 1.79%. 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